from notes written mostly by Dr. Carla Savage: All Rights Reserved

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1 CSC 505, Fall 2000: Week 9 Objectives: learn about various issues related to finding shortest paths in graphs learn algorithms for the single-source shortest-path problem observe the relationship among various algorithmically-induced spanning trees learn an algorithm for finding shortest paths among all pairs of vertices learn about the transitive closure of a graph Page 1 of 16

2 Shortest paths G = (V,E,w), directed or undirected. Path P = v 0 v 1...v k where v i v i+1 E for 0 i k 1. Weight of P = w(p ) = k 1 i=0 w(v i, v i+1 ) Shortest path weight from u to v (u, v V ): δ(u, v) = { min{w(p ) P is a path from u to v} if u v otherwise A shortest path from u to v is any path P from u to v with w(p ) = δ(u, v) Page 2 of 16

3 Shortest path problems Single pair: Given s, t V, find δ(s, t). Single source: Given s V, find δ(s, v) for all v V. All pairs: Find δ(u, v) for all u, v V. Potential Pitfalls Negative weight edges: s t Negative weight cycles: s t 9 1 Page 3 of 16

4 Optimal substructure property Lemma 9.1. If P = v 0 v 1...v k is ashortest path from v 0 to v k, then, for any i, j satisfying 0 i j k, Q = v i v i+1...v j is a shortest path from v i to v j. Proof. If not, suppose R is a shorter path from v i to v j. Q v i+1 v j 1 v 0 v 1 v v j v k 1 v k i R But if R isshorter than Q, then P = v 0...v i 1 R v j+1...v k is shorter than P. Contradiction. Page 4 of 16

5 Single-source shortest paths Source: s V No negative-weight cycles Initialize: for each v V do d[v] ; π[v] nil d[s] 0 d[v] is the shortest length of a path from s to v so far. π[v] isv s predecessor on that path. Invariant: d[v] δ(s, v) for all v V. do Relax(u,v) until d[v] =δ(s, v) for all v V procedure Relax(u, v) is if d[u]+w(u, v) <d[v] then d[v] d[u] +w(u, v); π[v] u end Relax Relax preserves the invariant. s u v Page 5 of 16

6 Dijkstra s algorithm Source: s V No negative weights Initialize; S Sis the set of vertices to which a shortest path is known. Add all vertices to a priority queue Q with key d[v] while Q is not empty do u Extract-Min(Q); S S {u} for each v on adj[u] do if v S then Relax(u, v); Decrease-Key(Q, v, d[v]) Claim: When u is added to S, d[u] =δ(s, u). Page 6 of 16

7 Example of Dijkstra s algorithm s x 1 a y 1 z b Page 7 of 16

8 Correctness of Dijkstra s algorithm Lemma 9.2. When u is added to S in Dijkstra s algorithm, d[u] = δ(s, u). Proof. By induction: the lemma is trivially true for s. Assume it to be true for all vertices added to S before u. Let P be a shortest path from s to u, let y be the first vertex of P not in S, and let x be y s predecessor on P. Let P(i, j) denote the portion of P from vertex i to vertex j. Then w(p) = w(p (s, x)) + w(x, y) + w(p(y,u)) w(p (s, x)) + w(x, y) no negative weights = d[x] + w(x, y) by induction d[y] Relax(x, y) was done d[u] u was chosen Thus δ(s, u) d[u] (by the above argument) and d[u] δ(s, u) (by the invariant on all SSSP algorithms). Page 8 of 16

9 Algorithmic spanning trees s S frontier vertices: those with neighbors in the set S Basic step is to choose a new frontier vertex (and the edge associated with it) DFS -- edges to frontier are on a stack BFS -- edges to frontier are on a queue (finds least number of hops). PFS -- edges to frontier are on a priority queue (at most one per vertex); Prim s (MST): key = w(e); Dijkstra s (SSSP): key = w(p(s,v)) Page 9 of 16

10 Bellman-Ford versus Dijkstra Bellman-Ford (SSSP with no restriction on edge weights): 1. Initialize d and π arrays, set d[s] =0. 2. Relax(u, v) for every uv E (repeat n 1 times). 3. If any edge uv has d[v] < d[u]+w(u, v), then report a negative cycle. Else d[u] = δ(s, u) for all u V. Dijkstra s best time is Θ(m + n lg n) Bellman-Ford s is Θ(mn). Bellman-Ford does not work on an undirected graph with negative-weight edges (such edges represent negative-weight cycles of length 2). Page 10 of 16

11 All-pairs shortest-path problem n repetitions of Dijkstra gives Θ(mn + n 2 lg n) when there are no negative-weight edges. n repetitions of Bellman-Ford gives Θ(mn 2 ) otherwise. Can we do better? All-pairs shortest-path problem on graph G = (V,E,w) with vertices labelled 1,..., n: Given an n n weight matrix W = {w ij }, where w ij = 0 if i = j, if ij E, and w(i, j) otherwise, Compute an n n distance matrix D = {d ij }, where d ij = δ(i, j) and an n n routing matrix R = {r ij }, where r ij is the vertex afer i on the shortest path from i to j. Page 11 of 16

12 Dynamic programming approach Failed divide-and-conquer approach (divide the solution all paths in half): D(k) = distances that use paths with at most k edges. d ij (k) = min ( ) d ij (k/2), min {d il(k/2) + d lj (k/2)} 1 l n Dynamic programming: Compute D(1) = W, D(2), D(4),..., D(2 p ), where p lg(n 1). Each entry takes Θ(n), each matrix takes Θ(n 3 ), and the total time is Θ(n 3 lg n). Page 12 of 16

13 Floyd-Warshall algorithm Another attempt: D(k) = distances using paths with intermediate vertices in {1,..., k}. D(0) = W D(k) : d ij (k) = min {d ij (k 1), d ik (k 1) + d kj (k 1)} i shortest path from i to j that uses 1,, k i doesn t use k => same as with k 1 j j i k does use k => composed of two paths that don t j Page 13 of 16

14 Implementation, time, and storage compute D = D(0) and R = R(0) for i 1 to n, j 1 to n do d ij w ij ; if w ij < then r ij j for k 1 to n do compute D = D(k) and R = R(k) for i 1 to n, j 1 to n do if d ik + d kj < d ij then d ij d ik + d kj ; r ij r ik Page 14 of 16

15 Transitive closure of a graph Suppose G = (V,E) is a directed (unweighted) graph. Then the transitive closure of G is G = (V,E ), where uv E iff u v in G. Given an n n boolean adjacency matrix A = {a ij }, where a ij = 1ifij E, and 0 otherwise, Compute the n n transitive closure matrix of A, X = {x ij }, where x ij = 1iffij E. same as Floyd-Warshall, except we only care whether or not a path exists for i 1 to n, j 1 to n do x ij a ij ; if i = j then x ij 1 for k 1 to n do for i 1 to n, j 1 to n do x ij x ij (x ik x kj ) Page 15 of 16

16 Relationship to matrix multiplication row i product (using boolean ops) is 1 iff there is a k so that i leads to k and k leads to j column j Can prove that there exists a Θ( f (n)) algorithm for transitive closure iff there exists a Θ( f (n)) algorithm for boolean matrix multiplication. [e.g. AHU, 1974] Page 16 of 16